Python Program For Pointing Arbit Pointer To Greatest Value Right Side Node In A Linked List

Given singly linked list with every node having an additional “arbitrary” pointer that currently points to NULL. We need to make the “arbitrary” pointer to the greatest value node in a linked list on its right side.

A Simple Solution is to traverse all nodes one by one. For every node, find the node which has the greatest value on the right side and change the next pointer.

Below is the implementation of the above approach:

Python3

# Python Program to point arbit pointers
# to highest value on its right

# Node class
class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None
        self.arbit = None

# Function to reverse the linked list
def reverse(head):
    prev = None
    current = head
    next = None

    while (current != None):
        next = current.next
        current.next = prev
        prev = current
        current = next

    return prev

# This function populates arbit pointer
# in every node to the greatest value
# to its right.
def populateArbit(head):

    # Reverse given linked list
    head = reverse(head)

    # Initialize pointer to maximum value node
    max = head

    # Traverse the reversed list
    temp = head.next
    while (temp != None):

        # Connect max through arbit pointer
        temp.arbit = max

        # Update max if required
        if (max.data < temp.data):
            max = temp

        # Move ahead in reversed list
        temp = temp.next

    # Reverse modified linked list and return head.
    return reverse(head)

# Utility function to print result linked list
def printNextArbitPointers(node):

    print("Node ", "Next Pointer " , "Arbit Pointer")
    while (node != None):

        print(node.data , "     ", end = "")

        if (node.next != None):
            print(node.next.data , "     ", end = "")
        else :
            print("NULL" , "     ",end = "")

        if (node.arbit != None):
            print(node.arbit.data, end = "")
        else :
            print("NULL", end = "")

        print("")
        node = node.next

# Function to create a new node with given data
def newNode(data):

    new_node = Node(0)
    new_node.data = data
    new_node.next = None
    return new_node

# Driver code
head = newNode(5)
head.next = newNode(10)
head.next.next = newNode(2)
head.next.next.next = newNode(3)
head = populateArbit(head)

print("Resultant Linked List is: ")
printNextArbitPointers(head)

# This code is contributed by Susobhan Akhuli

Output

Resultant Linked List is: 
Node  Next Pointer  Arbit Pointer
5      10      10
10      2      3
2      3      3
3      NULL      NULL

The Time Complexity of this above solution is O(n²).

Space complexity:
The space complexity of the program is O(1) as it uses a constant amount of extra space irrespective of the input size.

An Efficient Solution can work in O(n) time. Below are the steps.  

1. Reverse the given linked list.
2. Start traversing the linked list and store the maximum value node encountered so far. Make arbit of every node to point to max. If the data in the current node is more than the max node so far, update max.
3. Reverse modified linked list and return head.

Following is the implementation of the above steps.  

Python

# Python Program to point arbit pointers
# to highest value on its right

# Node class
class Node:

    # Constructor to initialize the
    # node object
    def __init__(self, data):
        self.data = data
        self.next = None
        self.arbit = None

# Function to reverse the linked list
def reverse(head):
    prev = None
    current = head
    next = None

    while (current != None):
        next = current.next
        current.next = prev
        prev = current
        current = next

    return prev

# This function populates arbit pointer
# in every node to the greatest value
# to its right.
def populateArbit(head):

    # Reverse given linked list
    head = reverse(head)

    # Initialize pointer to maximum
    # value node
    max = head

    # Traverse the reversed list
    temp = head.next
    while (temp != None):

        # Connect max through arbit
        # pointer
        temp.arbit = max

        # Update max if required
        if (max.data < temp.data):
            max = temp

        # Move ahead in reversed list
        temp = temp.next

    # Reverse modified linked list and
    # return head.
    return reverse(head)

# Utility function to print result
# linked list
def printNextArbitPointers(node):

    print("Node ",
        "Next Pointer " ,
        "Arbit Pointer")
    while (node != None):

        print(node.data ,
            "     ",
            end = "")

        if (node.next != None):
            print(node.next.data ,
                "     ", end = "")
        else :
            print("NULL" ,
                "     ",end = "")

        if (node.arbit != None):
            print(node.arbit.data, end = "")
        else :
            print("NULL", end = "")

        print("")
        node = node.next

# Function to create a new node
# with given data
def newNode(data):

    new_node = Node(0)
    new_node.data = data
    new_node.next = None
    return new_node

# Driver code
head = newNode(5)
head.next = newNode(10)
head.next.next = newNode(2)
head.next.next.next = newNode(3)
head = populateArbit(head)

print("Resultant Linked List is: ")
printNextArbitPointers(head)

# This code is contributed by Arnab Kundu
# This code is modified by Susobhan Akhuli

Output

Resultant Linked List is: 
Node  Next Pointer  Arbit Pointer
5      10      10
10      2      3
2      3      3
3      NULL      NULL

The time complexity of this program is O(n), where n is the number of nodes in the linked list. 

The space complexity of this program is also O(n), as it creates a new node for each element in the list and stores them in memory.

Recursive Solution:
We can recursively reach the last node and traverse the linked list from the end. The recursive solution doesn’t require reversing of the linked list. We can also use a stack in place of recursion to temporarily hold nodes. Thanks to Santosh Kumar Mishra for providing this solution. 

Python3

# Python3 program to point arbit pointers
# to highest value on its right

''' Link list node '''
# Node class
class newNode:

    # Constructor to initialize the
    # node object
    def __init__(self, data):
        self.data = data
        self.next = None
        self.arbit = None

# This function populates arbit pointer
# in every node to the greatest value
# to its right.
maxNode = newNode(None)
def populateArbit(head):

    # using static maxNode to keep track
    # of maximum orbit node address on
    # right side
    global maxNode

    # if head is null simply return the list
    if (head == None):
        return

    ''' if head.next is null it means we
        reached at the last node just update
        the max and maxNode '''
    if (head.next == None):
        maxNode = head
        return

    ''' Calling the populateArbit to the
        next node '''
    populateArbit(head.next)

    ''' updating the arbit node of the
        current node with the maximum
        value on the right side '''
    head.arbit = maxNode

    ''' if current Node value id greater
        then the previous right node then
        update it '''
    if (head.data > maxNode.data and
        maxNode.data != None ):
        maxNode = head
    return

# Utility function to prresult
# linked list
def printNextArbitPointers(node):
    print("Node    " ,
          "Next Pointer    " ,
          "Arbit Pointer")
    while (node != None):
        print(node.data,
              "        ",
              end = "")
        if(node.next):
            print(node.next.data,
                  "        ",
                  end = "")
        else:
            print("NULL", "        ",
                   end = "")
        if(node.arbit):
            print(node.arbit.data, end = "")
        else:
            print("NULL", end = "")
        print()
        node = node.next

# Driver code
head = newNode(5)
head.next = newNode(10)
head.next.next = newNode(2)
head.next.next.next = newNode(3)
populateArbit(head)
print("Resultant Linked List is:")
printNextArbitPointers(head)
# This code is contributed by SHUBHAMSINGH10

Output

Resultant Linked List is:
Node     Next Pointer     Arbit Pointer
5         10         10
10         2         3
2         3         3
3         NULL         NULL

Please refer complete article on Point arbit pointer to greatest value right side node in a linked list for more details!